Question

A rod of mass M, length L. Perpendicular force F at one end. Find its angular acceleration

Answer 1

\frac{3F}{ML}

Answer 2

The rod is assumed to be pivoted at one end. A perpendicular force F is applied at the other end. The torque about the pivot is $\tau = FL$. The moment of inertia of a uniform rod about one end is $I = \frac{1}{3}ML^2$. Using the rotational dynamics equation $\tau = I\alpha$, the angular acceleration $\alpha$ is calculated as $\alpha = \frac{\tau}{I} = \frac{FL}{\frac{1}{3}ML^2} = \frac{3F}{ML}$.